Advanced Calculus Single Variable

4.9 Cauchy Sequences And Completeness

You recall the definition of completeness which stated that every nonempty set of real
numbers which is bounded above has a least upper bound and that every nonempty set of real
numbers which is bounded below has a greatest lower bound and this is a property of the real
line known as the completeness axiom. Geometrically, this involved filling in the holes. There
is another way of describing completeness in terms of Cauchy sequences which will be
discussed soon.

Definition 4.9.1

{a }
n

is a Cauchy sequence if for all ε > 0, there exists nεsuchthat whenever n,m ≥ nε,

|an − am| < ε.

A sequence is Cauchy means the terms are “bunching up to each other” as m,n get
large.

Theorem 4.9.2The set of terms (values) of a Cauchy sequence in F isbounded.

Proof: Let ε = 1 in the definition of a Cauchy sequence and let n > n1. Then from the
definition,

|an − an1| < 1.

It follows that for all n > n1,

|an| < 1+ |an1|.

Therefore, for all n,

n1
|an| ≤ 1 + |an |+ ∑ |ak|. ■
1 k=1

Theorem 4.9.3If a sequence

{an}

in F converges, then the sequence is aCauchy sequence.

Proof: Let ε > 0 be given and suppose an→ a. Then from the definition of convergence,
there exists nε such that if n > nε, it follows that

ε
|an − a| < 2

Therefore, if m,n ≥ nε + 1, it follows that

ε ε
|an − am | ≤ |an − a|+ |a− am | < 2 + 2 = ε

showing that, since ε > 0 is arbitrary,

{an}

is a Cauchy sequence. ■

The following theorem is very useful.

Theorem 4.9.4Suppose

{an}

is a Cauchy sequence in F and thereexists asubsequence,

{ank}

which converges to a. Then

{an}

also converges to a.

Proof: Let ε > 0 be given. There exists N such that if m,n > N, then

|am − an| < ε∕2.

Also there exists K such that if k > K, then

|a− ank| < ε∕2.

Then let k > max

(K, N )

. Then for such k,

|ak − a| ≤ |ak − ank|+ |ank − a|
< ε∕2+ ε∕2 = ε.■

The next definition has to do with sequences which are real numbers.

Definition 4.9.5The sequenceof real numbers,

{an}

, is monotone increasingif for all n,an≤ an+1. The sequence ismonotone decreasing if for all n, an≥ an+1.People often leave off the word “monotone”.

If someone says a sequence is monotone, it usually means monotone increasing.

There exist different descriptions of completeness. An important result is the following
theorem which gives a version of completeness in terms of Cauchy sequences. This is often
more convenient to use than the earlier definition in terms of least upper bounds and greatest
lower bounds because this version of completeness, although it is equivalent to the
completeness axiom for the real line, also makes sense in many situations where Definition
2.10.1 on Page 61 does not make sense, ℂ for example because by Problem 12 on Page 86
there is no way to place an order on ℂ. This is also the case whenever the sequence is of
points in multiple dimensions.

It is the concept of completeness and the notion of limits which sets analysis apart from
algebra. You will find that every existence theorem in analysis depends on the assumption
that some space is complete.

Theorem 4.9.6Every Cauchysequence in ℝ converges if and only if everynon-empty set of real numbers which is bounded above has a least upper bound and everynonempty set of real numbers which is bounded below has a greatest lower bound.

Proof: First suppose every Cauchy sequence converges and let S be a nonempty set which
is bounded above. Let b1 be an upper bound. Pick s1∈ S. If s1 = b1, the least
upper bound has been found and equals b1. If

(s + b)
1 1

∕2 is an upper bound to S,
let this equal b2. If not, there exists b1> s2>

(s + b)
1 1

∕2 so let b2 = b1 and s2
be as just described. Now let b2 and s2 play the same role as s1 and b1 and do
the same argument. This yields a sequence

{s }
n

of points of S which is monotone
increasing and another sequence of upper bounds,

{b }
n

which is monotone decreasing
and

−n+1
|sn − bn| ≤ 2 (b1 − s1)

Therefore, if m > n

|bn − bm | ≤ bn − sm ≤ bn − sn ≤ 2−n+1 (b1 − s1)

and so

{bn}

is a Cauchy sequence. Therefore, it converges to some number b. Then b must be
an upper bound of S because if not, there would exist s > b and then

bn − b ≥ s− b

which would prevent

{bn}

from converging to b. The claim that every nonempty set of
numbers bounded below has a greatest lower bound follows similarly. Alternatively, you could
consider −S ≡

{− x : x ∈ S}

and apply what was just shown.

Now suppose the condition about existence of least upper bounds and greatest lower
bounds. Let

is a monotone increasing sequence which is bounded above, it convergesand whenever

{bn}

is a monotone sequence which is bounded below, it converges.

Proof: Let a = sup

{an : n ≥ 1}

and let ε > 0 be given. Then from Proposition 2.10.3 on
Page 62 there exists m such that a−ε < am≤ a. Since the sequence is increasing, it follows
that for all n ≥ m, a−ε < an≤ a. Thus a = limn→∞an. The case of a decreasing sequence is
similar. Alternatively, you could consider the sequence

{− an}

and apply what was just shown
to this decreasing sequence. ■

By Theorem 4.9.6 the following definition of completeness is equivalent to the original
definition when both apply.

Definition 4.9.8Whenever every Cauchy sequence in some set converges, theset is called complete.